I am having some problems when trying to format the time string I have created, I am trying to make it output only the time in HH:mm format by using the Date and Time conversion characters I found at this website
DateFormat time = new SimpleDateFormat("HH:mm R");
I get no problems without the "R" but then it outputs the entire date and time, defeating my goal.
Well, as R is not documented to mean anything but the documentation clearly says all other formatting characters "...from 'A' to 'Z' and from 'a' to 'z' are reserved...", all bets are off about what you'll get if you use R in the formatting string.
Remove the R. If your goal is to have an R at the end of the string, append it to the result. If not, consult the documentation to see what formatting character you should be using.
There is no "R" in the javadoc for SimpleDateFormat -- and it says that all unspecified letters are reserved. I don't know where your website got it...
You don't need 'R'. If you want 24 hour "military" time, just use capital 'HH' and be done with it. Delete the R and you should get what you want. If you don't, you're probably doing something else wrong in your code that you haven't posted.
According to the date/time formatting symbols for SimpleDateFormat, there is no "R". In fact, I get an IllegalArgumentException trying to use the "R" as you have it.
You can't use DateFormat time = new SimpleDateFormat("HH:mm R");.
However, printf recognizes a different set of symbols for formatting dates, and "R" is defined there:
'R' Time formatted for the 24-hour clock as "%tH:%tM"
The website you refer to does mention "R" -- under the printf section, not for SimpleDateFormat. It mentions to prepend t before all date/time format symbols.
Try
System.out.printf("%tR", new Date());
Related
I'm trying to parse the date format used in PDFs. According to this page, the format looks as follows:
D:YYYYMMDDHHmmSSOHH'mm'
Where all components except the year are optional. I assume this means the string can be cut off at any point as i.e. specifying a year and an hour without specifying a month and a day seems kind of pointless to me. Also, it would make parsing pretty much impossible.
As far as I can tell, Java does not support zone offsets containing single quotes. Therefore, the first step would be to get rid of those:
D:YYYYMMDDHHmmSSOHHmm
The resulting Java date pattern should then look like this:
['D:']uuuu[MM[dd[HH[mm[ss[X]]]]]]
And my overall code looks like this:
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("['D:']uuuu[MM[dd[HH[mm[ss[X]]]]]]");
TemporalAccessor temporalAccessor = formatter.parseBest("D:20020101",
ZonedDateTime::from,
LocalDateTime::from,
LocalDate::from,
Month::from,
Year::from
);
I would expect that to result in a LocalDate object but what I get is java.time.format.DateTimeParseException: Text 'D:20020101' could not be parsed at index 2.
I've played around a bit with that and found out that everything works fine with the optional literal at the beginning but as soon as I add optional date components, I get an exception.
Can anybody tell me what I'm doing wrong?
Thanks in advance!
I've found a solution:
String dateString = "D:20020101120000+01'00'";
String normalized = dateString.replace("'", "");
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("['D:']ppppy[ppM[ppd[ppH[ppm[pps[X]]]]]]");
TemporalAccessor temporalAccessor = formatter.parseBest(normalized,
OffsetDateTime::from,
LocalDateTime::from,
LocalDate::from,
YearMonth::from,
Year::from
);
As it seems, the length of the components is ambiguous and parsing of the date without any separators thus failed.
When specifying a padding, the length of each component is clearly stated and the date can therefore be parsed.
At least that's my theory.
I have found that SimpleDateFormat::parse(String source)'s behavior is (unfortunatelly) defaultly set as lenient: setLenient(true).
By default, parsing is lenient: If the input is not in the form used by this object's format method but can still be parsed as a date, then the parse succeeds.
If I set the leniency to false, the documentation said that with strict parsing, inputs must match this object's format. I have used paring with SimpleDateFormat without the lenient mode and by mistake, I had a typo in the date (letter o instead of number 0). (Here is the brief working code:)
// PASSED (year 199)
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd.mm.yyyy");
System.out.println(simpleDateFormat.parse("03.12.199o"));
simpleDateFormat.setLenient(false);
System.out.println(simpleDateFormat.parse("03.12.199o")); //WTF?
In my surprise, this has passed and no ParseException has been thrown. I'd go further:
// PASSED (year 1990)
String string = "just a String to mess with SimpleDateFormat";
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd.mm.yyyy");
System.out.println(simpleDateFormat.parse("03.12.1990" + string));
simpleDateFormat.setLenient(false);
System.out.println(simpleDateFormat.parse("03.12.1990" + string));
Let's go on:
// FAILED on the 2nd line
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd.mm.yyyy");
System.out.println(simpleDateFormat.parse("o3.12.1990"));
simpleDateFormat.setLenient(false);
System.out.println(simpleDateFormat.parse("o3.12.1990"));
Finally, the exception is thrown: Unparseable date: "o3.12.1990". I wonder where is the difference in the leniency and why the last line of my first code snippet has not thrown an exception? The documentation says:
With strict parsing, inputs must match this object's format.
My input clearly doesn't strictly match the format - I expect this parsing to be really strict. Why does this (not) happen?
Why does this (not) happen?
It’s not very well explained in the documentation.
With lenient parsing, the parser may use heuristics to interpret
inputs that do not precisely match this object's format. With strict
parsing, inputs must match this object's format.
The documentation does help a bit, though, by mentioning that it is the Calendar object that the DateFormat uses that is lenient. That Calendar object is not used for the parsing itself, but for interpreting the parsed values into a date and time (I am quoting DateFormat documentation since SimpleDateFormat is a subclass of DateFormat).
SimpleDateFormat, no matter if lenient or not, will accept 3-digit year, for example 199, even though you have specified yyyy in the format pattern string. The documentation says about year:
For parsing, if the number of pattern letters is more than 2, the year
is interpreted literally, regardless of the number of digits. So using
the pattern "MM/dd/yyyy", "01/11/12" parses to Jan 11, 12 A.D.
DateFormat, no matter if lenient or not, accepts and ignores text after the parsed text, like the small letter o in your first example. It objects to unexpected text before or inside the text, as when in your last example you put the letter o in front. The documentation of DateFormat.parse says:
The method may not use the entire text of the given string.
As I indirectly said, leniency makes a difference when interpreting the parsed values into a date and time. So a lenient SimpleDateFormat will interpret 29.02.2019 as 01.03.2019 because there are only 28 days in February 2019. A strict SimpleDateFormat will refuse to do that and will throw an exception. The default lenient behaviour can lead to very surprising and downright inexplicable results. As a simple example, giving the day, month and year in the wrong order: 1990.03.12 will result in August 11 year 17 AD (2001 years ago).
The solution
VGR already in a comment mentioned LocalDate from java.time, the modern Java date and time API. In my experience java.time is so much nicer to work with than the old date and time classes, so let’s give it a shot. Try a correct date string first:
DateTimeFormatter dateFormatter = DateTimeFormatter.ofPattern("dd.mm.yyyy");
System.out.println(LocalDate.parse("03.12.1990", dateFormatter));
We get:
java.time.format.DateTimeParseException: Text '03.12.1990' could not
be parsed: Unable to obtain LocalDate from TemporalAccessor:
{Year=1990, DayOfMonth=3, MinuteOfHour=12},ISO of type
java.time.format.Parsed
This is because I used your format pattern string of dd.mm.yyyy, where lowercase mm means minute. When we read the error message closely enough, it does state that the DateTimeFormatter interpreted 12 as minute of hour, which was not what we intended. While SimpleDateFormat tacitly accepted this (even when strict), java.time is more helpful in pointing out our mistake. What the message only indirectly says is that it is missing a month value. We need to use uppercase MM for month. At the same time I am trying your date string with the typo:
DateTimeFormatter dateFormatter = DateTimeFormatter.ofPattern("dd.MM.yyyy");
System.out.println(LocalDate.parse("03.12.199o", dateFormatter));
We get:
java.time.format.DateTimeParseException: Text '03.12.199o' could not
be parsed at index 6
Index 6 is where is says 199. It objects because we had specified 4 digits and are only supplying 3. The docs say:
The count of letters determines the minimum field width …
It would also object to unparsed text after the date. In short it seems to me that it gives you everything that you had expected.
Links
DateFormat.setLenient documentation
Oracle tutorial: Date Time explaining how to use java.time.
Leniency is not about whether the entire input matches but whether the format matches. Your input can still be 3.12.1990somecrap and it would work.
The actual parsing is done in parse(String, ParsePosition) which you could use as well. Basically parse(String) will pass a ParsePosition that is set up to start at index 0 and when the parsing is done the current index of that position is checked.
If it's still 0 the start of the input didn't match the format, not even in lenient mode.
However, to the parser 03.12.199 is a valid date and hence it stops at index 8 - which isn't 0 and thus the parsing succeeded. If you want to check whether everything was parsed you'd have to pass your own ParsePosition and check whether the index is matches to the length of the input.
If you use setLenient(false) it will still parse the date till the desired pattern is meet. However, it will check the output date is a valid date or not. In your case, 03.12.199 is a valid date, so it will not throw an exception. Lets take an example to understand where the setLenient(false) different from setLenient(true)/default.
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd.MM.yyyy");
System.out.println(simpleDateFormat.parse("31.02.2018"));
The above will give me output: Sat Mar 03 00:00:00 IST 2018
But the below code throw ParseException as 31.02.2018 is not a valid/possible date:
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd.MM.yyyy");
simpleDateFormat.setLenient(false);
System.out.println(simpleDateFormat.parse("31.02.2018"));
I want to define a pattern for the Java SimpleDaterFormat to parse existing strings.
The existing dates look like this: 2011-05-02T13:40:00+02:00.
I tried with different patterns, but I got ParseExceptions. The problem seems to be the timezone format.
Printing the pattern in Java:
yyyy-MM-dd'T'HH:mm:ssZ
2012-03-14T15:40:44+0100
yyyy-MM-dd'T'HH:mm:ssz
2012-03-14T15:41:58MEZ
But how can I get
???
2011-05-02T13:40:00+02:00
I'm using Java 6, not Java 7.
If you can use Java 7 or newer, you can use the XXX pattern to get the timezone to look like +02:00:
yyyy-MM-dd'T'HH:mm:ssXXX
Otherwise you might have to manipulate the date string to remove the colon from the timezone before parsing it.
I know it's a bit old question, but someone else might benefit from my hint.
You can use JodaTime. As library documentation stands:
Zone: 'Z' outputs offset without a colon, 'ZZ' outputs the offset with
a colon, 'ZZZ' or more outputs the zone id.
You can use it as well with java 6. You have more examples in this question
String dateString = "20110706 1607";
DateTimeFormatter dateStringFormat = DateTimeFormat.forPattern("YYYYMMDD HHMM");
DateTime dateTime = dateStringFormat.parseDateTime(dateString);
Resulting stacktrace:
Exception in thread "main" java.lang.IllegalArgumentException: Invalid format: "201107206 1607" is malformed at " 1607"
at org.joda.time.format.DateTimeFormatter.parseMillis(DateTimeFormatter.java:644)
at org.joda.time.convert.StringConverter.getInstantMillis(StringConverter.java:65)
at org.joda.time.base.BaseDateTime.<init>(BaseDateTime.java:171)
at org.joda.time.DateTime.<init>(DateTime.java:168)
......
Any thoughts? If I truncate the string to 20110706 with pattern "YYYYMMDD" it works, but I need the hour and minute values as well. What's odd is that I can convert a Jodatime DateTime to a String using the same pattern "YYYYMMDD HHMM" without issue
Thanks for looking
Look at your pattern - you're specifying "MM" twice. That can't possibly be right. That would be trying to parse the same field (month in this case) twice from two different bits of the text. Which would you expect to win? You want:
DateTimeFormat.forPattern("yyyyMMdd HHmm")
Look at the documentation for DateTimeFormat to see what everything means.
Note that although calling toString with that pattern will produce a string, it won't produce the string you want it to. I wouldn't be surprised if the output even included "YYYY" and "DD" due to the casing, although I can't test it right now. At the very least you'd have the month twice instead of the minutes appearing at the end.
Try this:
DateFormat df = new SimpleDateFormat("y");
System.out.println(df.format(new Date()));
Without reading the javadoc for SimpleDateFormat, what would you expect this to output? My expectation was "0". That is to say, the last digit of the current year, which is 2010.
Instead it pads it out to 2 digits, just as if the format string had been "yy".
Why? Seems rather bizarre. If I had wanted 2 digits then I'd have used "yy".
"Without reading the javadoc" is a dangerous attitude more often than not. Joshua Bloch wrote a book on this strange behaviours in Java.
That's why you need to check the documentation:
For parsing with the abbreviated year pattern ("y" or "yy"), SimpleDateFormat must interpret the abbreviated year relative to some century.
"y" is an abbreviation, so this is expected behavior.
If you want only a digit you can:
System.out.println(df.format(new Date()).substring(1));
:)